zoukankan      html  css  js  c++  java
  • POJ 3281 Dining 网络流最大流

    B - Dining
    Time Limit: 20 Sec

    Memory Limit: 256 MB

    题目连接

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88038#problem/B

    Description

    Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others.

    Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although he might not be able to stuff everybody, he wants to give a complete meal of both food and drink to as many cows as possible.

    Farmer John has cooked F (1 ≤ F ≤ 100) types of foods and prepared D (1 ≤ D ≤ 100) types of drinks. Each of his N (1 ≤ N ≤ 100) cows has decided whether she is willing to eat a particular food or drink a particular drink. Farmer John must assign a food type and a drink type to each cow to maximize the number of cows who get both.

    Each dish or drink can only be consumed by one cow (i.e., once food type 2 is assigned to a cow, no other cow can be assigned food type 2).

    Input

    Line 1: Three space-separated integers: NF, and D
    Lines 2.. N+1: Each line i starts with a two integers Fi and Di, the number of dishes that cow i likes and the number of drinks that cow i likes. The next Fiintegers denote the dishes that cow i will eat, and the Di integers following that denote the drinks that cow i will drink.

    Output

    Line 1: A single integer that is the maximum number of cows that can be fed both food and drink that conform to their wishes

    Sample Input

    4 3 3
    2 2 1 2 3 1
    2 2 2 3 1 2
    2 2 1 3 1 2
    2 1 1 3 3

    Sample Output

    3

    HINT

    题意

    有n头牛,每头牛有f种喜欢的食物和d种喜欢的饮料,然后问你,究竟多有多少头牛,可以吃到满意的食物和饮料

    题解

    s-食物-牛-牛-饮料

    因为牛只能用一次,所以注意拆点

    代码:

    //qscqesze
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    #include <map>
    #include <stack>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define maxn 5000
    #define mod 10007
    #define eps 1e-9
    int Num;
    //const int inf=0x7fffffff;   //нчоч╢С
    const int inf=0x3f3f3f3f;
    inline ll read()
    {
        ll x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    //**************************************************************************************
    namespace NetFlow
    {
        const int MAXN=100000,MAXM=100000,inf=1e9;
        struct Edge
        {
            int v,c,f,nx;
            Edge() {}
            Edge(int v,int c,int f,int nx):v(v),c(c),f(f),nx(nx) {}
        } E[MAXM];
        int G[MAXN],cur[MAXN],pre[MAXN],dis[MAXN],gap[MAXN],N,sz;
        void init(int _n)
        {
            N=_n,sz=0; memset(G,-1,sizeof(G[0])*N);
        }
        void link(int u,int v,int c)
        {
            E[sz]=Edge(v,c,0,G[u]); G[u]=sz++;
            E[sz]=Edge(u,0,0,G[v]); G[v]=sz++;
        }
        int ISAP(int S,int T)
        {//S -> T
            int maxflow=0,aug=inf,flag=false,u,v;
            for (int i=0;i<N;++i)cur[i]=G[i],gap[i]=dis[i]=0;
            for (gap[S]=N,u=pre[S]=S;dis[S]<N;flag=false)
            {
                for (int &it=cur[u];~it;it=E[it].nx)
                {
                    if (E[it].c>E[it].f&&dis[u]==dis[v=E[it].v]+1)
                    {
                        if (aug>E[it].c-E[it].f) aug=E[it].c-E[it].f;
                        pre[v]=u,u=v; flag=true;
                        if (u==T)
                        {
                            for (maxflow+=aug;u!=S;)
                            {
                                E[cur[u=pre[u]]].f+=aug;
                                E[cur[u]^1].f-=aug;
                            }
                            aug=inf;
                        }
                        break;
                    }
                }
                if (flag) continue;
                int mx=N;
                for (int it=G[u];~it;it=E[it].nx)
                {
                    if (E[it].c>E[it].f&&dis[E[it].v]<mx)
                    {
                        mx=dis[E[it].v]; cur[u]=it;
                    }
                }
                if ((--gap[dis[u]])==0) break;
                ++gap[dis[u]=mx+1]; u=pre[u];
            }
            return maxflow;
        }
        bool bfs(int S,int T)
        {
            static int Q[MAXN]; memset(dis,-1,sizeof(dis[0])*N);
            dis[S]=0; Q[0]=S;
            for (int h=0,t=1,u,v,it;h<t;++h)
            {
                for (u=Q[h],it=G[u];~it;it=E[it].nx)
                {
                    if (dis[v=E[it].v]==-1&&E[it].c>E[it].f)
                    {
                        dis[v]=dis[u]+1; Q[t++]=v;
                    }
                }
            }
            return dis[T]!=-1;
        }
        int dfs(int u,int T,int low)
        {
            if (u==T) return low;
            int ret=0,tmp,v;
            for (int &it=cur[u];~it&&ret<low;it=E[it].nx)
            {
                if (dis[v=E[it].v]==dis[u]+1&&E[it].c>E[it].f)
                {
                    if (tmp=dfs(v,T,min(low-ret,E[it].c-E[it].f)))
                    {
                        ret+=tmp; E[it].f+=tmp; E[it^1].f-=tmp;
                    }
                }
            }
            if (!ret) dis[u]=-1; return ret;
        }
        int dinic(int S,int T)
        {
            int maxflow=0,tmp;
            while (bfs(S,T))
            {
                memcpy(cur,G,sizeof(G[0])*N);
                while (tmp=dfs(S,T,inf)) maxflow+=tmp;
            }
            return maxflow;
        }
    }
    using namespace NetFlow;
    map<pair<int,int> ,int> H;
    int tot=1;
    int get_id(int x,int y)
    {
        pair<int,int> A;
        A.first = x;
        A.second = y;
        if(H[A]!=0)
            return H[A];
        else H[A]=tot++;
        return H[A];
    }
    
    int main()
    {
        init(10000);
        int n=read(),f=read(),d=read();
        for(int i=1;i<=f;i++)
            link(get_id(1,0),get_id(i,2),1);
        for(int i=1;i<=d;i++)
            link(get_id(i,3),get_id(2,0),1);
        for(int i=1;i<=n;i++)
            link(get_id(i,1),get_id(i,4),1);
        for(int i=1;i<=n;i++)
        {
            int ff=read(),dd=read();
            for(int j=1;j<=ff;j++)
            {
                int x=read();
                link(get_id(x,2),get_id(i,1),1);
            }
            for(int j=1;j<=dd;j++)
            {
                int x=read();
                link(get_id(i,4),get_id(x,3),1);
            }
        }
        printf("%d
    ",dinic(get_id(1,0),get_id(2,0)));
    }
  • 相关阅读:
    Size Classes with Xcode 6:为所有的尺寸准备一个Storyboard
    iOS:界面适配(二)--iPhone/iPad适配(关于xib)
    iPhone屏幕尺寸、分辨率及适配
    ViewControl的size设为freeform
    iphone分辨率大全
    iPhone6分辨率与适配
    Applications using Launch Screen Files and targetting iOS 7.1 and earlier need to also include a Launch Image in an Asset Catalog.
    linuxmint计算器
    校验 MD5 值
    DataBinding
  • 原文地址:https://www.cnblogs.com/qscqesze/p/4733379.html
Copyright © 2011-2022 走看看